?an?(1)求证:数列??是等差数列;
?n??1?(2)求数列??的前n项和Sn.
?an?
【答案】(1)见解析;(2)Sn?n.
2?n?1?2【解析】(1)nan?1??n?1?an?2n?2n的两边同时除以n?n?1?,得
an?1an??2?n?N??, n?1n?an?∴数列??是首项为4,公差为2的等差数列
?n?a(2)由(1),得n?2n?2,
n∴an?2n2?2n,故
111?n?1??n1?11??2???????, an2n?2n2n?n?1?2?nn?1?∴Sn?1??1??11?1???11?????????????? 2??nn?1????2??23??1??111??111??1?1?n1???????????1????????2?n??23n?1????23?2?n?1?2?n?1?.
9
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